The energy released in the following fission reaction is approximately:
\( {}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{142}_{56}\mathrm{Ba} + {}^{91}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n} + Q \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 200\,\mathrm{MeV} \)
The energy released in fission comes from the increase in binding energy per nucleon: a heavy nucleus such as \( {}^{235}\mathrm{U} \) (about 7.6 MeV per nucleon) splits into middle-mass fragments (about 8.5 MeV per nucleon).
\( \text{Gain per nucleon} \approx 8.5 - 7.6 = 0.9\,\mathrm{MeV} \)
\( Q \approx 236\times0.9\,\mathrm{MeV} \approx 200\,\mathrm{MeV} \)
This is the standard figure of about 200 MeV per fission of uranium-235, used when no masses are given.
NCERT (2026-27):
Fission – “… The energy released (the Q value ) in the fission reaction of nuclei like uranium is of the order of 200 MeV per fissioning nucleus.”